Spectral Theorem Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice the spectral theorem: recognizing real symmetric and complex Hermitian matrices, proving eigenvalues are real, using orthogonality of eigenspaces, building \(A=QDQ^T\) or \(A=UDU^*\), reading \(\operatorname{tr}A\), \(\det A\), rank, and powers from eigenvalues, expanding \(A=\sum_i\lambda_i q_iq_i^T\), classifying quadratic forms by eigenvalue signs, and spotting projection matrices with eigenvalues \(0\) and \(1\). Open the lesson for focused worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this spectral theorem practice works
1. Take the practice set: answer questions about symmetric matrices, Hermitian matrices, orthogonal diagonalization, spectral decompositions, trace, determinant, rank, powers, Rayleigh quotients, and definiteness.
2. Open the lesson: review the theorem, recognition tests, worked examples, and single-answer checks.
3. Retry: return to the question set and first decide whether the problem asks about symmetry, eigenvectors, diagonal form, spectral data, or a quadratic form.
What you will learn in the spectral theorem lesson
Self-adjoint matrices
Real case: \(A^T=A\) is the signal for the real spectral theorem
Complex case: \(A^*=A\) is Hermitian and has real eigenvalues
Eigenspaces for distinct eigenvalues are orthogonal
Orthogonal diagonalization
Real symmetric matrices admit \(A=QDQ^T\) with \(Q^TQ=I\)
The columns of \(Q\) are an orthonormal eigenbasis
Repeated eigenvalues still allow orthonormal bases inside their eigenspaces
Spectral decomposition
Write \(A=\sum_i\lambda_i q_iq_i^T\) using rank-one orthogonal projections
Powers and functions act on eigenvalues: \(f(A)=Qf(D)Q^T\)
Trace, determinant, rank, and invertibility are read from eigenvalues
Quadratic forms and projections
Use \(x^TAx=\sum_i\lambda_i y_i^2\) after an orthonormal coordinate change
Positive definite means all eigenvalues are positive
Symmetric projections have eigenvalues only \(0\) and \(1\)
Purpose: Build a reliable workflow for spectral theorem problems: recognize self-adjoint matrices, use real eigenvalues and orthogonal eigenspaces, form \(A=QDQ^T\) or \(A=UDU^*\), interpret \(A=\sum_i\lambda_i q_iq_i^T\), and classify quadratic forms from eigenvalue signs.
Success criteria
Recognize real symmetric matrices by \(A^T=A\) and Hermitian matrices by \(A^*=A\).
State that self-adjoint matrices have real eigenvalues.
Use orthogonality of eigenspaces for distinct eigenvalues.
Build an orthonormal eigenbasis, including within repeated eigenspaces.
Read \(Q^{-1}=Q^T\) in \(A=QDQ^T\) and \(U^{-1}=U^*\) in \(A=UDU^*\).
Use spectral decomposition \(A=\sum_i\lambda_i q_iq_i^T\).
Compute trace, determinant, rank, powers, and invertibility from eigenvalues.
Classify definiteness of \(x^TAx\) from the signs of the eigenvalues.
Key vocabulary
Self-adjoint: equal to its adjoint; \(A^T=A\) over \(\mathbb{R}\), \(A^*=A\) over \(\mathbb{C}\).
Orthogonal matrix: \(Q^TQ=I\), so \(Q^{-1}=Q^T\).
Unitary matrix: \(U^*U=I\), so \(U^{-1}=U^*\).
Orthonormal eigenbasis: a basis of unit eigenvectors that are pairwise orthogonal.
Spectral projection: \(q_iq_i^T\), the rank-one projection onto the eigenline spanned by \(q_i\).
Rayleigh quotient: \(\dfrac{x^TAx}{x^Tx}\), bounded by the smallest and largest eigenvalues for symmetric \(A\).
Quick pre-check
Pre-check: The real spectral theorem applies most directly to which matrices?
Hint: The theorem needs the matrix to agree with its transpose in the real case.
Symmetric and Hermitian matrices have real spectral data
Learning goal: Know when the spectral theorem is available and what it guarantees before computing anything.
Key idea
In a real inner product space, the key condition is \(A^T=A\). In a complex inner product space, the matching condition is \(A^*=A\), where \(A^*\) is conjugate transpose. These self-adjoint conditions force eigenvalues to be real and make the operator behave like a stretch along perpendicular directions.
Recognition checklist
First check that the matrix is square.
Real case: compare entries across the diagonal, so \(a_{ij}=a_{ji}\).
Complex case: compare \(a_{ij}\) with \(\overline{a_{ji}}\).
If the condition holds, look for an orthonormal eigenbasis rather than a general change-of-basis matrix.
Worked example
Example: Let \(A=\operatorname{diag}(2,5)\). Why is the spectral theorem immediate?
The matrix is real symmetric because it equals its transpose. The standard basis vectors \(e_1,e_2\) are already orthonormal eigenvectors, with eigenvalues \(2\) and \(5\). Thus \(A=QDQ^T\) with \(Q=I\) and \(D=\operatorname{diag}(2,5)\).
Try it
Try it: Which matrix is symmetric?
Hint: A real symmetric matrix has matching entries above and below the diagonal.
Different eigenvalues give perpendicular eigenspaces
Learning goal: Use the proof idea that symmetry moves \(A\) across an inner product.
Key idea
If \(Au=\lambda u\) and \(Av=\mu v\), then symmetry gives \(\langle Au,v\rangle=\langle u,Av\rangle\). Therefore \(\lambda\langle u,v\rangle=\mu\langle u,v\rangle\). When \(\lambda≠\mu\), this forces \(\langle u,v\rangle=0\).
Repeated eigenvalues
Distinct eigenspaces are automatically orthogonal.
Inside one repeated-eigenvalue eigenspace, vectors are not automatically orthogonal.
Use Gram-Schmidt inside a repeated eigenspace to choose an orthonormal basis.
Combining these bases across all eigenspaces gives the orthonormal eigenbasis promised by the theorem.
Worked example
Example: The matrix \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) has eigenvectors \((1,1)\) and \((1,-1)\). What happens after normalization?
The eigenvalues are \(1\) and \(-1\). The eigenvectors have dot product \(1-1=0\), so they are orthogonal. After scaling by \(1/\sqrt2\), they become orthonormal columns of a matrix \(Q\).
Try it
Try it: For a real symmetric matrix, eigenvectors associated with distinct eigenvalues are what?
Hint: Use \(\langle Au,v\rangle=\langle u,Av\rangle\) and subtract the two scalar multiples.
Put eigenvectors into \(Q\), put eigenvalues into \(D\)
Learning goal: Translate an orthonormal eigenbasis into the diagonal form used in computations.
Key idea
For a real symmetric matrix, choose orthonormal eigenvectors \(q_1,\dots,q_n\). Let \(Q\) have these vectors as columns and let \(D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Then \(AQ=QD\), so \(A=QDQ^T\). In the complex Hermitian case, the form is \(A=UDU^*\).
Formula notes
The diagonal entries of \(D\) are eigenvalues, repeated with multiplicity.
The columns of \(Q\) are the matching unit eigenvectors.
Orthogonal means \(Q^TQ=I\), so \(Q^{-1}=Q^T\).
The order of eigenvectors may change the order of diagonal entries, but not the operator.
Use \(q_1=(1,1)/\sqrt2\) with eigenvalue \(1\) and \(q_2=(1,-1)/\sqrt2\) with eigenvalue \(-1\). Then \(Q=[q_1\ q_2]\) is orthogonal and \(D=\operatorname{diag}(1,-1)\), so \(A=QDQ^T\).
Try it
Try it: If \(Q\) is orthogonal, what is \(Q^{-1}\)?
Hint: Orthogonal columns satisfy \(Q^TQ=I\).
A symmetric matrix is a weighted sum of orthogonal projections
Learning goal: Read actions, powers, rank, trace, and determinant directly from eigenvalues.
Key idea
From \(A=QDQ^T\), with columns \(q_i\), expand the product to get \[A=\sum_i \lambda_i q_iq_i^T.\] Each \(q_iq_i^T\) projects onto one unit eigenline, and \(\lambda_i\) tells how strongly \(A\) stretches that direction.
Spectral data
\(\operatorname{tr}A=\sum_i\lambda_i\).
\(\det A=\prod_i\lambda_i\).
\(\operatorname{rank}A\) is the number of nonzero eigenvalues.
\(A^k=QD^kQ^T\), so eigenvalues of \(A^k\) are \(\lambda_i^k\).
If all \(\lambda_i≠0\), then \(A^{-1}=QD^{-1}Q^T\).
Worked example
Example: A real symmetric matrix has eigenvalues \(0,0,5\). What are its rank, determinant, and invertibility status?
Only one eigenvalue is nonzero, so the rank is \(1\). The determinant is the product \(0\cdot0\cdot5=0\), so the matrix is singular and not invertible.
Try it
Try it: If \(A\) is symmetric with eigenvalues \(2\) and \(3\), what are the eigenvalues of \(A^2\)?
Hint: Powers act on eigenvalues by the same power.
Eigenvalue signs classify \(x^TAx\)
Learning goal: Turn a quadratic form into a diagonal sum by rotating to an orthonormal eigenbasis.
Key idea
If \(A=QDQ^T\) and \(y=Q^Tx\), then \[x^TAx=y^TDy=\sum_i\lambda_i y_i^2.\] Because \(Q\) preserves lengths, the signs of the eigenvalues determine whether the quadratic form is positive, negative, semidefinite, or indefinite.
Definiteness tests
All eigenvalues positive: positive definite.
All eigenvalues nonnegative and at least one zero: positive semidefinite but not definite.
All eigenvalues negative: negative definite.
All eigenvalues nonpositive and at least one zero: negative semidefinite but not definite.
At least one positive and at least one negative eigenvalue: indefinite.
Worked example
Example: A real symmetric matrix has eigenvalues \(0,4\). How should its quadratic form be classified?
Both eigenvalues are nonnegative, and one eigenvalue is zero. Therefore \(x^TAx\ge0\) for every \(x\), but it is not positive for every nonzero \(x\). The form is positive semidefinite, not positive definite.
Try it
Try it: A real symmetric matrix has eigenvalues \(-1\) and \(3\). How is its quadratic form classified?
Hint: One eigenvalue gives a negative direction and the other gives a positive direction.
Projections and Rayleigh quotients are spectral examples
Learning goal: Connect the theorem to common operators that appear in linear algebra problems.
Key idea
A symmetric projection \(P\) satisfies \(P^2=P\) and \(P^T=P\). If \(Pv=\lambda v\), then \(P^2v=Pv\) gives \(\lambda^2=\lambda\), so \(\lambda\) is \(0\) or \(1\). The spectral theorem says the space splits orthogonally into the range and kernel of the projection.
Operator view
Symmetric projections have diagonal form with only \(0\) and \(1\) on the diagonal.
The trace of a symmetric projection equals its rank.
The Rayleigh quotient of symmetric \(A\) lies between the smallest and largest eigenvalues.
The operator norm of symmetric \(A\) is \(\max_i|\lambda_i|\).
Worked example
Example: What is the spectral picture of projection onto a line in \(\mathbb{R}^2\), and why does a unit eigenvector \(q\) with \(Aq=4q\) satisfy \(q^TAq=4\)?
The line direction is an eigenline with eigenvalue \(1\), because vectors on the line are unchanged. The perpendicular direction is an eigenline with eigenvalue \(0\), because it is sent to zero. In an orthonormal basis adapted to the line, the projection matrix is \(\operatorname{diag}(1,0)\). For the Rayleigh check, \(q^TAq=q^T(4q)=4q^Tq=4\) because \(\|q\|=1\).
Try it
Try it: A symmetric projection matrix is orthogonally diagonalizable with which possible diagonal entries?
Hint: Apply \(P^2=P\) to an eigenvector.
Most mistakes ignore the hypotheses or the basis
Learning goal: Finish with a compact checklist for common spectral theorem errors.
Common traps
Diagonalizable is not enough: the theorem needs self-adjoint structure for an orthonormal eigenbasis.
Symmetric vs. skew-symmetric: \(A^T=A\), not \(A^T=-A\).
Repeated eigenvalues: you may need to orthonormalize inside one eigenspace.
Order matching: each diagonal entry in \(D\) must match the corresponding column of \(Q\).
Definiteness: zeros give semidefinite, not definite.
Projection matrices: \(P^2=P\) gives a projection; \(P^T=P\) gives an orthogonal projection in Euclidean space.
Worked example
Example: If \(A=5I\) in \(\mathbb{R}^3\), how many eigenvectors are available?
Every nonzero vector is an eigenvector with eigenvalue \(5\). The repeated eigenspace is all of \(\mathbb{R}^3\), so choose any orthonormal basis. This is a good reminder that repeated eigenvalues do not prevent the spectral theorem.
Try it
Try it: A Hermitian matrix has eigenvalues that are what?
Hint: Hermitian matrices are the complex self-adjoint analogue of real symmetric matrices.
Final recap
Real symmetric means \(A^T=A\); Hermitian means \(A^*=A\).
Self-adjoint matrices have real eigenvalues.
Distinct eigenspaces are orthogonal.
Repeated eigenspaces can be given orthonormal bases.
Real symmetric matrices satisfy \(A=QDQ^T\) with \(Q\) orthogonal.
Spectral decomposition is \(A=\sum_i\lambda_i q_iq_i^T\).
Trace, determinant, rank, powers, and invertibility are read from eigenvalues.
For a unit eigenvector \(q\), the Rayleigh quotient gives \(q^TAq=\lambda\).
Quadratic form definiteness is controlled by eigenvalue signs.
Symmetric projections have eigenvalues \(0\) and \(1\).
Next step: Close this lesson and try the quiz again. For each problem, first check the matrix type, then decide whether the answer is about eigenvalue reality, orthogonality, diagonal form, spectral decomposition, trace, determinant, rank, a Rayleigh quotient, or quadratic-form signs.
Practice set
Twierdzenie spektralne practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Do których macierzy twierdzenie spektralne w postaci rzeczywistej odnosi się najbezpośredniej?
Correct answer: A. Rzeczywiste macierze symetryczne
Explanation: Rzeczywiste macierze symetryczne mają ortonormalną bazę wektorów własnych.
Question 2Not answered
Co można powiedzieć o wartościach własnych rzeczywistej macierzy symetrycznej?
Correct answer: D. Są rzeczywiste
Explanation: Rzeczywiste macierze symetryczne mają rzeczywiste wartości własne.
Question 3Not answered
Wektory własne macierzy symetrycznej odpowiadające różnym wartościom własnym są:
Correct answer: B. Prostopadłe
Explanation: Symetria wymusza, aby przestrzenie własne dla różnych wartości własnych były prostopadłe.
Question 4Not answered
Rzeczywista macierz symetryczna jest diagonalizowana przez:
Correct answer: A. Macierz ortogonalną
Explanation: Twierdzenie spektralne daje diagonalizację ortogonalną.
Question 5Not answered
Jeśli \(A=QDQ^T\) przy ortogonalnej \(Q\), to czym jest \(Q^{-1}\)?
Correct answer: B. \(Q^T\)
Explanation: Dla macierzy ortogonalnej macierz odwrotna jest równa jej transpozycji.
Question 6Not answered
Jeśli macierz symetryczna ma wartości własne \(1\) i \(3\), to jakie są elementy diagonalne jej spektralnej postaci diagonalnej?
Correct answer: B. \(1\) i \(3\)
Explanation: Postać diagonalna zawiera wartości własne.
Question 7Not answered
Która macierz jest symetryczna?
Correct answer: D. \(\begin{pmatrix}1&2\\2&3\end{pmatrix}\)
Explanation: Macierz symetryczna jest równa swojej transpozycji.
Question 8Not answered
Jeśli macierz symetryczna ma powtórzoną wartość własną, twierdzenie spektralne nadal daje:
Correct answer: C. Ortonormalną bazę wektorów własnych
Explanation: Nawet dla powtórzonych przestrzeni własnych można dobrać bazę ortonormalną.
Question 9Not answered
Jakie jest geometryczne znaczenie diagonalizacji ortogonalnej?
Correct answer: D. Istnieje ortonormalna baza wektorów własnych
Explanation: Oznacza to, że przekształcenie jest diagonalne w bazie ortonormalnej.
Question 10Not answered
Czy dla rzeczywistej macierzy symetrycznej gwarantowana jest diagonalizowalność?
Correct answer: B. Tak
Explanation: Tak. To jeden z głównych wniosków twierdzenia spektralnego.
